1. Warm Up Simplify the square roots

2. Sections 7.1-7.2 Pythagorean Theorem and its Converse
Objective:
To find missing sides of right triangles; to determine if a triangle is a right triangle.

3. Pythagorean Theorem Where a and b are the legs of the
right triangle and c is the hypotenuse. c a b

4. Use the Pythagorean Theorem to solve for the variable
Use the Pythagorean Theorem to solve for the variable. Write your answer in reduced radical form and decimal form. 8 y 17
2.) 15 36 x
1.)
82 + y2 = 172
= x2
64 + y2 = 289
= x2
y2 = 225
1521 = x2
y =
225
y = 15
39 = x

5. Pythagorean Triple
A set of three positive integers a, b, and c that satisfy the equation
Pythagorean Triples:
3, 4, 5
5, 12, 13
8, 15, 17
7, 24, 25
And any multiple of the list above (p.435)

6. The given lengths are two sides of a right triangle
The given lengths are two sides of a right triangle. All three sides of the triangle form a Pythagorean Triple. Find the length of the third side and tell whether it is a leg or hypotenuse.
1.) 40 and ) 12 and ) 63 and 65
122 = 144
632 = 3969
402 = 1600
352 = 1225
652 = 4225
412 = 1681
Sum is 1369, which is 372
Difference is 256, which is 162
Difference is 81, which is 92
Third side is 9. It’s a leg.
Third side is 37. It’s a hypotenuse.
Third side is 16. It’s a leg.

7. Converse of Pythagorean Theorem
: Right Triangle
: Acute Triangle
: Obtuse Triangle

8. Decide whether the numbers represent the side lengths of a triangle
Decide whether the numbers represent the side lengths of a triangle. If they can, classify the triangles as right, acute, or obtuse.
1.) 20, 21, ) 15, 36, ) 14, 48, 50
20+21>28
21+28>20
20+28>21
15+36>39
36+39>15
15+39>36
14+48>50
48+50>14
14+50>48
282 ?
392 ?
502 ?
784 ?
1521 ?
2500 ?
784 < 841
1521 =1521
2500 =2500
acute
right
right

9. Area of a triangle
Area = ½ bh, where the base and height are perpendicular to each other. h h b b

10. Find the area of the triangle given A = ½ bh
12 cm
10 ft.
7 ft.
8 cm
122 = a2 + 42
144 = a2 + 16
128 = a2
A = 35 ft2
A = 45.2 cm2
11.3 = a

11. Find the unknown side length.
2.) x
1.) x 2
= x2 +
X2 + X2 = ( )2
2x2 = 64(2)
19 = x2 + 7
2x2 = 128
12 = x2
x2 = 64
x =
x = 8

12. Homework Textbook pages 430-432 #3-25 every other odd
Textbook page 438 #3-25 every other odd